标签：a long next numbers 不能 call ali int 用两个 one
描述
Recently, my lovely daughter YuYu indulges in playing a game called dominoes. When the game starts, I first place some blocks with varying heights in a long row. I told her that she can select only one of them and topple it to fall either left or right. If the falling block hits another one, it will cause it to fall over too (if the top of a falling block just touches the bottom of another one, that block still stays standing). YuYu is so gready that she wants to knock over as many blocks as possible. Now, your task is to write a program and help YuYu to find the maximum number of blocks she can knock over.
输入
There will be multiple test cases. Each test case starts with an integer n indicating the number of blocks.
The next line contains n(1<=n<=10000) pairs of integers. Each pair, x h, indicates the position x of a block of height h (x>=0, h>0 and all numbers will be no larger than 32767).
You must help YuYu to select the block to initially knock over (and in what direction) so as to knock over as many blocks as possible.
Input will terminate with an integer 0 on a line.
输出
For each input test case, output a line containing an integer m indicating the maximum number of blocks that can be knocked over.
样例输入
2
0 2 3 3
3
1 2 2 3 3 2
0
样例输出
1
3
// 可以跨着推倒 1 1 3 1 5 5 的话虽然3 1 不能推倒 1 1 但是5 5可以所以结果应该是3
所以只要现在牌的高度=max（前一个牌的高度减去前一个牌到现在牌的距离），（现在牌的高度）就可以了因为有正推和反推所以用两个数组避免影响。
#include<bits/stdc++.h> using namespace std; struct jilu { int x,h; }node[10001],node1[10001]; bool cmp(jilu a,jilu b) { return a.x<b.x; } int main() { std::ios::sync_with_stdio(false); int n; while(cin>>n) { if(n==0)break; int max1=1; for(int i=0;i<n;i++) { cin>>node[i].x>>node[i].h; node1[i].x=node[i].x; node1[i].h=node[i].h; } sort(node,node+n,cmp); sort(node1,node1+n,cmp); int sum=1; for(int i=1;i<n;i++) { if(node[i].x-node[i-1].x<node[i-1].h) { sum++; node[i].h=max(node[i-1].h-(node[i].x-node[i-1].x),node[i].h); } else sum=1; if(sum>max1) max1=sum; } sum=1; for(int i=n-1;i>0;i--) { if(node1[i].x-node1[i-1].x<node1[i].h) { sum++; node1[i-1].h=max(node1[i].h-(node1[i].x-node1[i-1].x),node1[i-1].h); } else sum=1; if(sum>max1) max1=sum; } cout<<max1<<endl; } }
标签：a long next numbers 不能 call ali int 用两个 one
原文地址：https://www.cnblogs.com/xbqdsjh/p/11563491.html
